11. Evaluate
- The sum of first three terms of a G.P. is 39/10 and their product is 1. Find the common ratio and the terms.
Solution:
Let a/r, a, ar be the first three terms of the G.P.
a/r + a + ar = 39/10 —-(1)
(a/r)(a)(ar) = 1 —–(2)
From (2), we obtain a3 = 1
a = 1
Substuting a = 1 in equation (1), we get
1/r + 1 + r = 39/10
1 + r + r2 = 39/10 r
10 + 10r + 10r2 – 39r = 0
10r2 – 29r + 10 = 0
10r2 – 25r – 4r + 10 = 0
5r(2r – 5)-2(2r – 5)= 0
(5r – 2)(2r – 5) = 0
r = 2/5 or 5/2
Thus the three terms of G.P. are 5/2, 1 and 2/5
- How many terms of G.P. 3, 32,33 … are needed to give the sum 120?
Solution:
The given G.P. is 3, 32, 33,…
Let n terms of this G.P. be required to obtain the sum as 120.
Thus, four terms of the given G.P. are required to obtain the sum as 120.
14: The sum of first three terms of a G.P. is 16 and the sum of the next three terms is 128. Determine the first term, the common ratio and the sum of n terms of the G.P.
Solution:
Let the G.P. be a, ar, ar2,…
According to the given condition,
a + ar + ar2 = 16 and ar3 + ar4 + ar5 = 128
⇒ a (1 + r + r2) = 16 ……………. (1)
ar3(1 + r + r2) = 128 …………….. (2)
Dividing equation (2) by (1), we obtain
- Given a G.P. with a = 729 and 7th term 64, determine S7
Solution:
a = 729 and a7= 64
Let r be the common ratio of the G.P. It is known that
an = a.rn-1
a7 = a.r7-1 =(729)r6
64 = 729(r6)
r6 = 64/729
r6 =(2/3)6
r = 2/3
Also it is known that,
- Find a G.P. for which sum of the first two terms is -4 and the fifth term is 4 times the third term.
Solution:
Let a be the first term and r be the common ratio of the G.P.
According to the given conditions,
Thus the required G.P. is –4/3, –8/3, –16/3,… or 4, -8, 16, -32,…
- If the 4th ,10th and 16th terms of a G.P. are x , y and z respectively. Prove that x , y, z are in G.P.
Solution:
Let a be the first term and r be the common ratio of the G.P.
According to the given condition,
a4 = ar3 = x ———-(1)
a10 = ar9 = y ———-(2)
a16 = ar15 = z ———-(3)
Dividing (2) by (1), we get
y/x = (ar9)/(ar3)
y/x = r6
Dividing (3) by (2), we get
z/y = (ar15)/(ar9)
z/y = r6
y/x = z/y
Thus x, y, z are in G.P.
- Find the sum to n terms of the sequence, 8, 88, 888, 8888…
Solution:
The given sequence is 8, 88, 888, 8888…
This sequence is not a G.P. However, it can be changed to G.P. as
Sn = 8 + 88 + 888 + 8888 + …………….. to n terms
= 8/9 [9 +99 + 999+ 9999+ … to n terms]
= 8/9 [(10-1) +(100-1) + (1000-1) + (1000-1) + … to n terms]
= 8/9 [(10 +102 + 103 + … to n terms)-(1+1+1+…n terms)]
= 8/9 [10(10^n-1)/9 – n]
= 80/81 (10n – 1) – 8/9 n
- Find the sum of the products of the corresponding terms of the sequences 2, 4, 8, 16, 32 and 128, 32, 8, 2, 1⁄2 .
Solution:
Required sum = 2×128 + 4×32 + 8×8 +16×2 +32x1⁄2
=64[4+2+1+1/2 +1/22]
Here 4, 2, 1, 1/2, 1/22 is a G.P.
First term a = 4
Common ratio r = 1/2
Required sum = 64(31/4) = 16(31) = 496
- Show that the products of the corresponding terms of the sequnnces form a, ar, ar2,…arn-1 and A, AR,AR2,…,ARn-1 a G.P. and find the common ratio.
Solution:
It has to be proved that the sequence: aA, arAR, ar2AR2,…arn-1ARn-1 forms a G.P.
Second term/first term = arAR/aA = rR
Third term /secondterm = ar2AR2/arAR = rR
Thus the above sequence forms a G.P. and the common ratio is rR
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